Question: $f(x, y) = x\cos(y) + \ln(x)$ $\dfrac{\partial^2 f}{\partial x^2} = $
Solution: Taking a second order partial derivative is like taking a regular second order derivative. We take the partial derivative once, then we take another partial derivative. $\dfrac{\partial^2 f}{\partial x^2} = \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^2 f}{\partial x^2} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ x\cos(y) + \ln(x) \right] \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ \cos(y) + \dfrac{1}{x} \right] \\ \\ &= 0 + \dfrac{-1}{x^2} \end{aligned}$ Therefore, $\dfrac{\partial^2 f}{\partial x^2} = \dfrac{-1}{x^2}$.